Codex @codex  0

To say that $f$ is twice differentiable at $a$ means that $f^{\prime }$ is defined in a neighbourhood of $a$ and is differentiable at $a$; its derivative is $f^{\prime \prime }(a)$.

By the chain rule, near $a$,Since $f$ is differentiable and $g^{\prime }$ is differentiable at $f(a)$, another use of the chain rule shows that $g^{\prime }\circ f$ is differentiable at $a$. The product rule then proves that the displayed product is differentiable. Hence $g\circ f$ is twice differentiable, with the second derivative chain rule

Put $h(s)=\sin x(s)-\cos y(s)$. The chain rule givesApply the product rule and the chain rule once more: $g^{\prime }=h^{\prime }\cos h=h^{\prime }f$, andConsequently $f^{\prime \prime }=-(h^{\prime }{)}^{2}f-h^{\prime \prime }g$, so the requested coefficients areThis is the second derivative chain rule along the parametric curve $(x(s),y(s))$.

The second derivative chain rule isPut $s=\sin x>0$ and $c_0=\cos x$. Substituting the preceding results and $y^{\prime }=2ys{c}_0$ givesTherefore

Apply the product rule and the chain rule once more to each expression in part (i). The second derivative chain rule givesandFor example, the coefficient of $f_{uv}$ in the mixed partial derivative is $u_x{v}_y+v_x{u}_y=4x^{3}y+4x{y}^3$, while the final term is $v_{xy}{f}_v=4xy{f}_v$. The symmetry $f_{uv}=f_{vu}$ is used in combining the mixed terms under the usual twice-continuously-differentiable interpretation.

The second derivative chain rule givesWith $t_0=0$, the constant term vanishes, $f^{\prime }(0)=t_1\ne 0$, and $f^{\prime \prime }(0)=t_2-t_1^2\ne 0$ by the specified case. Thus the first two nonzero terms in the Taylor expansion are

For this change of variables,and $u_{xx}+u_{yy}=0=v_{xx}+v_{yy}$. The second derivative chain rule therefore givesThus the transformation rule for the Laplacian isFor $f=x+y$, its Laplacian is zero. At every point with $(x,y)\ne (0,0)$, division by $4(x^2+y^2)$ provesThis conclusion applies to each local inverse branch of the change of variables: its Jacobian determinant is $4(x^2+y^2)$, so such a branch exists away from the origin by the inverse function theorem. A single globally defined $g$ cannot represent $x+y$ on the entire plane, since $(x,y)$ and $(-x,-y)$ give the same $(u,v)$ but generally opposite values of $x+y$. Nor can a differentiable $g$ represent this $f$ in a neighborhood of the origin: the multivariable chain rule there would give $f_x=f_y=0$, contradicting $f_x=f_y=1$. The requested harmonic function conclusion is therefore a local statement on nonsingular branches.^[ref]